silvertab wins

dialectics!

:glugglug

What r u talking about??

There is no ABC..... one has already gone.... he has opened a window and either A, B or C has gone

So the question only has 2 answers...... A or B

What is 3 - 1 ?

U have 3 possible answers as in the origional question.... take away one.... what is left?

2 answers...... 50/50

U have 2 options

sorry...you don't understand the problem

let's say you pick A.....

if you STAY with your choice, you will only win if the car is behind door A...

if you switch, you will win no matter if the car is behind B or C...TRY IT damn it

Ok here's the deal....

NOW EVERYONE WHO DOUBT IT JUST READ THIS

We will assume that there is a winning door and that the two remaining doors, A and B, both have goats behind them. There are three options:

1. The contestant first chooses the door with the car behind it. She is then shown either door A or door B, which reveals a goat. If she changes her choice of doors, she loses. If she stays with her original choice, she wins.

2. The contestant first chooses door A. She is then shown door B, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.

3. The contestant first chooses door B. She is then is shown door A, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.



Just fucking try it on your OWN.....

If your first pick is A...
The only time you will by NOT SWITCHING is if it's really behind door A...

Now if the the car is behind eighter B or C...you WILL WIN BY SWITCHING...IN BOTH FUCKING CASE

FUUUUUUUUUUUUUUUUUUUUUUUUCK

Trippy. I played the first link with the Java app. It still makes no sense to me that when the first option is removed it's not a 50/50 anymore, but by the java application it did prove that always switching ended in more wins than not.

So yea, if I'm ever on a game show and I get this message, I'll plug your name for ya. Even though I still only see two choices and not three.

Damn! Finally someone who decided to try it on his own! ;)

Thanks for being reasonable....

Still makes no sense. But I also don't have a degree in fluid dynamics, but I can usually drink without spilling too.... (huh ?)

here's a better place to test it

http://www-2.cs.cmu.edu/~donna/personal/MontyHall/


be sure to check
Monty always pick losing door


it's pretty easy when you think about it psili:

let's say you pick A....the only way you will win by not switching is if it's really behind A....but if it's behind B, Monty will show door C...you switch to B, you win!...if it's behind door C...monty will open door B, you switch to C, you win

So assuming you pick door A...and you don't switch, you will win 1/3 of the time (if it's behind door A)...if you pick A and you SWITCH...you will win, whater it's behind B or C... (2/3 of the time)

I can't believe you're still trying to explain it to them... :)

WTF>>>> I AM LOST>>>>> I WANT THE ORIGIONAL QUESTION POSTED HERE!!

here's the exact question

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?

which came first - the chicken or the egg?

well, that's just what I asked in other words

Trust the numbers man. I still don't quite get how the numbers come out like they do, but "always switching" always results in higher net wins than "never switching".

OK I get it....

U assume the host is crooked

I am a trusting idiot and I assume people are honest

very good :thumbsup

silvertab drew an awesome picture explaining it

hahaha well that was the question in the first place ;)

you guys are either really patient or very bored.

it's pretty simple, really. 66%.

when picking randomly out of 3 boxes, you ALWAYS have a 1:3 chance of winning = which means you are going to be wrong 2:3. Eliminate 1 of the wrong answers AFTER you have taken, with 66% probability, the other wrong answer - and you have your explanation.

But if I were predisposed to throwing away an empty box, me throwing away a single box would simply mean that _if it's one of my two boxes, it isn't that one_. That doesn't lower the chance of it being one of my two boxes though.

One should not forget the math, because math yields truth.

Your comparison, on the other hand, does not. You fail to account for the fact that the box the contestant chose has immunity from being revealed because you chose it. Thus, the fact that it goes on to the next round is meaningless.


Let's compare it to the models again:
10 models get voted for in a live tv show - all at once - and afterwards you get to pick one of them as your personal favorite.

Now, after the vote, one by one, the models with the least amount of votes are sent off. However, the model you chose as your favorite can stay until the final two no matter how many votes she got.

Clearly, in the final round, she'll have way less chance of winning than the other model, simply because there's an 8/10 chance that your model is only there because you chose her. The fact that you having chosen her gave her immunity until the last round has no impact whatsoever on the amount of votes she got.

So, by default, the 8/10 chance that she's just there because of the immunity you gave her goes to the other girl, and only within the remaining 2/10 chance does she have 1/2 chance - so, what she actually has is 1/10 chance.


In the case of the 3 boxes, the immunity acounts for 1/3 (which by default goes to the box which did not have immunity), and the initial decision has 1/2 of the remaining 2/3 - or a total of exactly 1/3 chance.

See, this is where i think the problem is. You say "The host reveals B or C"

How can two seperate events be counted as one?

Nobody has adequately explained to me why that is.

I've gone through every possible combination of events in this problem and i wind up with a 50% chance of winning.

Outcome number 1:

Prize is in Box A.
Contestant chooses Box A.
Box C is removed.
Stays = Win
Switches = Lose

Outcome number 2:

Prize is in Box A.
Contestant chooses Box A.
Box B is removed.
Stays = Win
Switches = Lose

Outcome number 3:

Prize is in Box A.
Contestant chooses Box B.
Box C is removed.
Stays = Loses
Switches = Win

Outcome number 4:

Prize is in Box A.
Contestant chooses Box C.
Box B is removed.
Stays = Loses
Switches = Win

Nobody has explained to me why the first two outcomes together are somehow only as likely as one of the other outcomes?

Equinox kept posting the results from that C program that someone had written. I can't read programming for shit. My suspicion is that the person has also lumped the first 2 outcomes into a 1 in 3 chance instead of a 2 in 4 chance.

In the end, there are 4 possible outcomes regardless of what box is initially chosen. And out of those 4 outcomes, 2 will win when you stay, and 2 will win when you switch.

Okay, i've thought about it some more. Initially, you have a 2 in 3 chance of choosing a situation that will lead to switching winning, whereas you only have a 1 in 3 chance of choosing a situation where staying will win.

I'M CONVERTED!

So what punkworld said about the two staying choices being a subset is correct. They each become a 1 in 6 chance.

In my defence, it was 6 am when we were discussing this earlier. :winkwink:

haha.

Nice! all this work was not in vain! ;)

I do not like this

as you will only be winning 5$ per winning session...

unless you have alot of money, and alot of time, to be begging 50, 100, 200, 400, 800 and such